I'm doing a course in probability and was asked the prove the following. I'd appreciate some feedback on it, thank you.
Let $X$ be a self independent r.v. . Prove that exists $c\in \mathbb{R}$ s.t. $\mathbb{P}(X=c)=1$
My proof:
Let A denote the support of $X$.
Lets assume that exists $x,y\in A$ such that $x\neq y$.
Let $B_1= \{x\}$ and $B_2=\{y\}$.
So we get: $\mathbb{P}(X\in B_1 \cap B_2) = \mathbb{P}(X\in B_1, \;X\in B_2) = \mathbb{P}(X\in B_1) \cdot \mathbb{P}(X\in B_2)=q_1 \cdot q_2 >0$
(this is true because $x,y \in A$ and have a non zero probability and $X$ is self independent).
But $B_1 \cap B_2 = \emptyset \Rightarrow \mathbb{P}(X \in B_1 \cap B_2)=0$ - Contradiction!
And since the support of a r.v. can not be empty we get that $A=\{x\}$ and by definition
$\mathbb{P}(X=x)=1$
Edit: Thank you for your comments, I'll fix it
By "self independent" I presume you mean $X$ and $X$ are independent.
Suggestion: Rather than looking at $P(X = x)$ which might always be $0$, consider the cumulative distribution function $F(x) = P(X \le x)$. Show that this is always $0$ or $1$.